note that for ANY fractions a/b and c/d, saying a/b = c/d is THE SAME as saying: ad = bc.

so let's "cross-multiply" to eliminate the "fractions":

\(\displaystyle (\sin\theta - \cos\theta + 1)(\cos\theta) = (\sin\theta + \cos\theta - 1)(\sin\theta + 1)\)

a little algebra, now:

\(\displaystyle \sin\theta\cos\theta - \cos^2\theta + \cos\theta = \sin^2\theta + \sin\theta\cos\theta - \sin\theta + \sin\theta + \cos\theta - 1\)

\(\displaystyle \sin\theta\cos\theta - \cos^2\theta + \cos\theta = \sin^2\theta + \sin\theta\cos\theta + \cos\theta - 1\)

we have some terms that occur on both sides, so we can subtract them from each side:

\(\displaystyle -\cos^2\theta = \sin^2\theta - 1\)

(see it coming)? put the trig terms on one side, and the constant terms on the other:

\(\displaystyle 1 = \sin^2\theta + \cos^2\theta\).

now, to be really slick, write this proof "upside-down", starting with the basic trig identity, and ending with what you wanted to prove.